Question

Find the value of $x$ for which ${\sec }^{-1}x+{\sin }^{-1}x=\frac{\pi }{2}.$

Answer 1

Solution: given,

$\Rightarrow {\sec }^{-1}x+{\sin }^{-1}x=\frac{\pi }{2}$

We know ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

and ${\sec }^{-1}x={\cos }^{-1}\frac{1}{x}$

$\Rightarrow {\cos }^{-1}\frac{1}{x}+{\sin }^{-1}x=\frac{\pi }{2}$

$\Rightarrow \frac{\pi }{2}-{\sin }^{-1}\frac{1}{x}+{\sin }^{-1}x=\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}x-{\sin }^{-1}\frac{1}{x}=0$

$\Rightarrow {\sin }^{-1}(x\sqrt{1-\frac{1}{x^2}}-\frac{1}{x}\sqrt{1-x^2})=0$

$\Rightarrow x\times \frac{\sqrt{x^2-1}}{x}-\frac{1}{x}\sqrt{1-x^2}=\sin 0$

$\Rightarrow \sqrt{x^2-1}-\frac{1}{x}\sqrt{1-x^2}=0$

$\Rightarrow \sqrt{x^2-1}=\frac{\sqrt{1-x^2}}{x}$ squaring both side

$\Rightarrow (x^2-1)=\frac{1-x^2}{x^2}$

$\Rightarrow (x^2-1)=\frac{-(x^2-1)}{x^2}$

$\Rightarrow x^2=-1$

$\Rightarrow x=\pm i$ Ans (here $i^2=-1$)